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Tolerance stacks: when the parts all sit at their limits

Stack parts in a line and the nominals add, and so do the tolerances. Worst case is the plain sum; the statistical estimate is the root of the sum of squares.

Hone is a place to practise a career, one idea a day. This is one of its lessons, written out in full and free to read without an account.

What it is for

Three spacers each within tolerance, and the assembly will not go into its housing. Nobody added the tolerances. The customer receives the one assembly where every part happened to be at its top limit on the same day.

How to think about it

Add the nominals for the nominal stack. Add the tolerances for the worst case; that is the number a fit must survive. If there are many parts and they are made independently, the root-sum-square is a fair estimate of what you will usually see, but it is not a guarantee.

Worked example

Worst case: nominals add, and tolerances add
The rule that guarantees the fit.
Three spacers: 5 ± 0.05, 12 ± 0.1, 8 ± 0.1 mm
The parts.
Nominal = 25 mm; worst-case ± 0.25 mm, so 24.75 to 25.25 mm
Every part at its limit in the same direction.
Statistical (RSS): sqrt(0.05² + 0.1² + 0.1²) = sqrt(0.0225) = ± 0.15 mm
What you will usually see if the parts are many and independent. Not what you may promise.

Your turn

Parts 6 ± 0.1 and 9 ± 0.2. Write the worst-case tolerance.

± (0.1 + ) = ± 0.3 mm

The trap

Averaging the tolerances, or assuming they cancel. Worst case is the sum: every part can sit at its limit on the same day, and the assembly that does not go together is the one that ships.

Practise tolerance stacks on HoneA question on it now, a coding challenge where there is one, and it is remembered for review. Free, no email needed.